Integer squares of the form $x^4+x^3+x^2+x+1$

Question

I want to solve ${y^2}=x^4+x^3+x^2+x+1$ in $\mathbb{Z}$.

I can find four solutions. Is there another solution? I know that there are six solutions. $$(-1, \pm 1)\;,\;(0, \pm 1).$$ My try: \begin{align} & x^4+x^3+x^2+x+1=0\quad \Rightarrow \underbrace{\left( x+\frac{1}{x} \right)^2}_{t^2}+\underbrace{\left( x+\frac{1}{x} \right)}_{t}-1=0\,\Rightarrow t\notin \mathbb{Z} \\ & x^4+x^3+x^2+x+1={\pm 1}^2\quad\Rightarrow \,x=-1\,\,,\,x=0.\\ \end{align}

Thanks.